Hambleton et al. (1991) suggest using a chi-square test to identify local independence between two items. The procedure consists of constructing a 2x2 table for two items using the correct and incorrect answers of persons at the same level of ability (i.e., with the same raw scores on the test):
Item X | |||
Item Y | Right | Wrong | Total |
Right Wrong |
A C |
B D |
Y_{R} = A+B N-Y_{R}= C+D |
Total | X_{R} = A+C | N-X_{R} = B+D | N = A+B+C+D |
This yields a chi-square statistic with 1 degree of freedom (without Yates' continuity correction, but the principle is the same):
This is minimized to zero, and local independence is apparently assured in this dataset at this ability level for this pair of items, when
AD = BC
i.e., when
A/N = X_{R}/N * Y_{R}/N
as it would be when the data exactly conform to the Rasch model, or any other model in which the two items are conditionally independent.
In conventional usage, however, when χ^{2} < 3.84, there is a presumption of local independence with 95% confidence. Then "local independence" is declared!
Let us conduct an experiment based on this convention. Let us constrain the possible outcomes so that, if a subject succeeds on item X, that subject cannot fail on item Y. So the items are always locally dependent. The data matrix becomes:
Item X | |||
Item Y | Right | Wrong | Total |
Right Wrong |
A 0 |
B D |
Y_{R} = A+B N-Y_{R}= D |
Total | X_{R} = A | N-X_{R} = B+D | N = A+B+D |
The chi-square test apparently becomes:
Then we can determine the relationship between successes on items X and Y for any value of chi-square, say N*k, where k is a constant. Accordingly,
Rewriting this in terms of probabilities, where P_{X} is the probability of success on item X under these conditions, etc., and taking logarithms, such that K=log_{e}(k):
It is seen that the relationship between the items is expressed in terms of log-odds difficulties, in accord with the Rasch model, despite that fact that the data do not accord with Rasch model conditions (because the items are locally dependent). Item Y is always easier than item X.
Further, for any particular chi-square "significance" value, e.g., 3.84, items with paired log-odds difficulties differing by less than log_{e} (3.84 / N) might be unwittingly declared "locally independent".
Here is an example:
Item X | |||
Item Y | Right | Wrong | Total |
Right Wrong |
20 0 |
67 13 |
87 13 |
Total | 20 | 80 | 100 |
The log-odds difficulties of the two items are log_{e}(13/87) = -1.90, and log_{e}(80/20) = 1.39 logits, differing by -3.29. The criterion value is log_{e}(3.84/100) = -3.26. Since -3.29 is less than -3.26, the items might be misleadingly declared "locally independent". For comparison, the chi-square computation above gives: χ² = 100 * 260^2 / (87 * 13* 20 * 80) = 3.74 with 1 d.f. one-sided → p=.053, so that the null hypothesis of local independence is not rejected.
Agustin Tristan
Hambleton R.K., Swaminathan H., Rogers H.J. (1991) "Fundamentals of item response theory", Sage publications Inc. London, Chapter 2, pp 9-12 and Exercise 6, pp 29-31
Chi-square local independence meets the Rasch model. Tristan A. … 16:1 p.861
Chi-square local independence meets the Rasch model. Tristan A. … Rasch Measurement Transactions, 2002, 16:1 p.861
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